ln(x^2+1) / (x^2+1) の積分

Problem / 問題

Show that: / 以下を示せ：
$${\int }_0^{\mathrm{\infty }}\frac{\ln (x^2+1)}{x^2+1}dx=\pi \ln 2$$

Solution / 解答

Let $x=\tan \theta$, $dx={\sec }^2\theta d\theta$.

$${\int }_0^{\mathrm{\infty }}\frac{\ln (x^2+1)}{x^2+1}dx={\int }_0^{\pi /2}\frac{\ln ({\sec }^2\theta )}{{\sec }^2\theta }{\sec }^2\theta d\theta =-2{\int }_0^{\pi /2}\ln (\cos \theta )d\theta$$

Let $I={\int }_0^{\pi /2}\ln (\cos \theta )d\theta$. Then $\theta \to \frac{\pi }{2}-\theta$ gives $I={\int }_0^{\pi /2}\ln (\sin \theta )d\theta$.

Hence $$2I+\frac{\pi }{2}\ln 2={\int }_0^{\pi /2}(\ln (\cos \theta )+\ln (\sin \theta )+\ln 2)d\theta ={\int }_0^{\pi /2}\ln (\sin 2\theta )d\theta$$

By symmetry, ${\int }_0^{\pi /2}\ln (\sin 2\theta )d\theta =2{\int }_0^{\pi /4}\ln (\sin 2\theta )d\theta ={\int }_0^{\pi /2}\ln (\sin \varphi )d\varphi =I$, where $\varphi =2\theta$. Hence $2I+\frac{\pi }{2}\ln 2=I$; $I=-\frac{\pi }{2}\ln 2$; ${\int }_0^{\mathrm{\infty }}\frac{\ln (x^2+1)}{x^2+1}dx=\pi \ln 2$.